YTTERBIUM GROUND STATES
By Prof. L. Kaliambos (Natural Philosopher in New Energy) March 27, 2016 Ytterbium is an atom of the chemical element ytterbium with symbol Yb and atomic number 70. However unlike for hydrogen, a closed-form solution to the Schrödinger equation for the many-electron atoms like the ytterbium atom has not been found. So, under the invalid relativity (EXPERIMENTS REJECT RELATIVITY) various approximations, such as the Hartree–Fock method, could be used to estimate the ground state energies. Under these difficulties I analyzed carefully the electromagnetic interactions of two spinning electrons of opposite spin which give a simple formula for the solution of such ground state energies. You can see it in my paper “ Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures”.(2008). Hence the electronic structure of the ytterbium ground states should be given by this correct image as 1s2.2s2.2px2.2py2.2pz2.3s2.3px2.3py2.3pz2.3d10.4s2.4p6.4d10.5s2.5px2.5py2.5pz2.4f14 .6s2 According to the “Ionization energies of the elements-WIKIPEDIA" the ionization energies in eV of ytterbium (from (E1 to E4 ) are the following: E1 = 6.25 , E2 = 12.18 , E3 = 25 , and E4 = 43.56 . Note that in the absence of more ionization energies we cannot calculate the ground state energies of inner orbitals. Nevertheless for understanding better such ground state energies you can see also my papers about the ground state energies of elements in my FUNDAMENTAL PHYSICS CONCEPTS. Moreover in “User Kaliambos” you can see my paper of 2008. GROUND STATE ENERGY OF -'(E1 + E2) = - 18.43 eV = E(6s2)' Here the E(6s2) represents the ground state energy of 6s2. The electron charges (-68e) of the 68 electrons of the following electron configuration (1s22s22p63s23p63d104s24p64d105s25p64f14) screen the nuclear charge (+ 70e) and for a perfect screening we would have ζ = 2. However the electrons of 6s penetrate the electron cloud of 4f and lead to the deformation of electron clouds. Thus ζ > 2 . Note that the 6s2 consists of one pair (2 electrons of opposite spin). Thus applying my formula of 2008 we write -E(6s2) = 18.43 = - (16.95)ζ + 4.1 / n2 Then using n = 6 the above equation can be written as (27.21)ζ2 - (16.95)ζ - 659.38 = 0 and solving for ζ we get ζ = 5.24 > 2 . Of course the two electrons of opposite spin (6s2) do not provide any mutual repulsion, because I discovered in 2008 that at very short inter-electron separations the magnetic attraction is stronger than the electric repulsion giving a vibration energy, which seems to be like a simple electric repulsion of the Coulomb law. This situation of a vibration energy due to an electromagnetic interaction indeed occurs, because the peripheral velocity of a spinning electron is faster than the speed of light, which invalidates Einstein’s theory of special relativity. (See my FASTER THAN LIGHT). However under the influence of invalid relativity and in the absence of a detailed knowledge about the mutual electromagnetic interaction between the electrons of opposite spin today many physicists believe incorrectly that two electrons with opposite spin exert a mutual Coulomb repulsion. Under such fallacious ideas I published my paper of 2008 . ' EXPLANATION OF E3 = 25 eV = -E(4f2) + E(4f1)' According to the experiments the electrons of 6s2 and 4f14 belong to the same energy level with n = 6 . The 4f14consist of seven pairs ( 14 electrons of 4f2 , 4f2 , 4f2 , 4f2, 4f2 , 4f2 and 4f2 with opposite spin). Here the E(4f2) represents the binding energy of the first 4f2, while the E(4f1) represents the binding energy of 4f1 which appears after the E3 ionization. The charges (-54e) of the 54 electrons of the following electron configuration (1s22s22p63s23p63d104s24p64d105s25p6) screen the nuclear charge (+70e) and for a perfect screening we would have ζ = 16. However in the absence of 6s2the electrons of 4f penetrate strongly the 5p6 and provide an effective ζ , whose the value is less than the perfect screening with ζ = 16. Thus ζ < 16 . Note that the first 4f2 consists of one pair (2 electrons of opposite spin). Thus applying my formula of 2008 we write -E(4f2) = -+ (16.95)ζ - 4.1 / n2 On the other hand, since the 4f1 consists of one electron, we apply the Bohr formula as E(4f1) = (-13.6057)ζ2/n2 Therefore E3 = 25 eV = -E(4f2) + E(4f1) = - (16.95)ζ + 4.1) / n2 Then using n = 6 the above equation can be written as (13.6057)ζ2 - (16.95)ζ - 895.9 = 0 Then solving for ζ we get ζ = 8.76 < 16 . ' ' EXPLANATION OF E4 = 43.56 eV = -E(4f2) + E(4f1) Here the E(4f2) represents the binding energy of the second 4f2, while the E(4f1) represents the binding energy of 4f1,which appears after the ionization of E4 . As in the case of E3 the charges (-54e) of the 54 electrons of the following electron configuration (1s22s22p63s23p63d104s24p64d105s25p6) screen the nuclear charge (+70e) and for a perfect screening we would have ζ = 16. However in the absence of 6s2and after the third ionization the new electron cloud of 4f penetrates also strongly the 5p6 . Thus it provides an effective ζ, whose the value is less than the perfect screening with ζ = 16. Note that the second 4f2 consists of one pair (2 electrons of opposite spin). Thus applying my formula of 2008 we write -E(4f2) = -+ (16.95)ζ - 4.1 / n2 On the other hand, since the 4f1 consists of one electron, we apply the Bohr formula as E(4f1) = (-13.6057)ζ2/n2 Therefore E4 = 43.56 eV = -E(4f2) + E(4f1) = - (16.95)ζ + 4.1) / n2 Then using n = 6 the above equation can be written as (13.6057)ζ2 - (16.95)ζ - 1564.06 = 0 Then solving for ζ we get ζ = 11.36 < 16 . 'WHY EINSTEIN’S WRONG RELATIVITY LED TO THE ABANDONMENT OF NATURAL LAWS IN FAVOR OF WRONG ATOMIC THEORIES ' It is of interest to note that in the absence of a detailed knowledge about the electromagnetic interaction of two electrons of opposite spin physicists so far under the influence of Einstein’s relativity could not calculate such ground state energies of many-electron atoms. Historically, despite the enormous success of the Bohr model and the quantum mechanics of the Schrodinger equation based on the well-established laws of electromagnetism in explaining the principal features of the hydrogen spectrum and of other one-electron atomic systems, so far, under the influence of Einstein’s relativity which led to the abandonment of natural laws neither was able to provide a satisfactory explanation of the two-electron atoms. In atomic physics a two-electron atom is a quantum mechanical system consisting of one nucleus with a charge Ze and just two electrons. This is the first case of many-electron systems. The first few two-electron atoms are: Z =1 : H- hydrogen anion. Z = 2 : He helium atom. Z = 3 : Li+ lithium atom anion. Z = 4 : Be2+ beryllium ion. Prior to the development of quantum mechanics, an atom with many electrons was portrayed like the solar system, with the electrons representing the planets circulating about the nuclear “sun”. In the solar system, the gravitational interaction between planets is quite small compared with that between any planet and the very massive sun; interplanetary interactions can, therefore, be treated as small perturbations. However, In the helium atom with two electrons, the interaction energy between the two spinning electrons and between an electron and the nucleus are almost of the same magnitude, and a perturbation approach is inapplicable. In 1925 the two young Dutch physicists Uhlenbeck and Goudsmit discovered the electron spin according to which the peripheral velocity of a spinning electron is greater than the speed of light. Since this discovery invalidates Einstein’s relativity it met much opposition by physicists including Pauli. Under the influence of Einstein’s invalid relativity physicists believed that in nature cannot exist velocities faster than the speed of light. So, great physicists like Pauli, Heisenberg, and Dirac abandoned the natural laws of electromagnetism in favor of wrong theories including qualitative approaches under an idea of symmetry properties between the two electrons of opposite spin which lead to many complications. Thus, in the “Helium atom-Wikipedia” one reads: “Unlike for hydrogen a closed form solution to the Schrodinger equation for the helium atom has not been found. However various approximations such as the Hartree-Fock method ,can be used to estimate the ground state energy and wave function of atoms”. It is of interest to note that in 1993 in Olympia of Greece I presented at the international conference “Frontiers of fundamental physics” my paper of dipole photons. In that paper I showed that LAWS AND EXPERIMENTS INVALIDATE FIELDS AND RELATIVITY . At the same time I tried to find not only the nuclear force and structure but also the coupling of two electrons under the application of the abandoned electromagnetic laws. For example in the photoelectric effect the absorption of light contributeς not only to the increase of the electron energy but also to the increase of the electron mass, because the particles of light have mass m = hν/c2. ( See my paper "DISCOVERY OF PHOTON MASS" ). However the electron spin which gives a peripheral velocity greater than the speed of light cannot be affected by the photon absorption. Thus after 9 years I presented at the 12th symposium of the Hellenic nuclear physics society my paper “Nuclear structure is governed by the fundamental laws of electromagnetism". (NCSR “Demokritos, 2002). The paper was published in Ind. J. Th. Phys (2003) in which I showed not only my DISCOVERY OF NUCLEAR FORCE AND STRUCTURE but also that the peripheral velocity (u >> c) of two spinning electrons with opposite spin gives an attractive magnetic force (Fm) stronger than the electric repulsion (Fe) when the two electrons of mass m and charge (-e) are at a very short separation (r < 578.8 /1015 m). Because of the antiparallel spin along the radial direction the interaction of the electron charges gives an electromagnetic force Fem = Fe - Fm . Therefore in my research the integration for calculating the mutual Fem led to the following relation: Fem = Fe - Fm = Ke2/r2 - (Ke2/r4)(9h2/16π2m2c2) Of course for Fe = Fm one gets the equilibrium separation ro = 3h/4πmc = 578.8/1015 m. That is, for an interelectron separation r < 578.8/1015 m the two electrons of opposite spin exert an attractive electromagnetic force, because the attractive Fm is stronger than the repulsive Fe . Here Fm is a spin-dependent force of short range. As a consequence this situation provides the physical basis for understanding the pairing of two electrons described qualitatively by the Pauli principle, which cannot be applied in the simplest case of the deuteron in nuclear physics, because the binding energy between the two spinning nucleons occurs when the spin is not opposite (S=0) but parallel (S=1). According to the experiments in the case of two electrons with antiparallel spin the presence of a very strong external magnetic field gives parallel spin (S=1) with electric and magnetic repulsions given by Fem = Fe + Fm So, according to the well-established laws of electromagnetism after a detailed analysis of paired electrons in two-electron atoms I concluded that at r < 578.8/1015 m a motional EMF produces vibrations of paired electrons. Unfortunately today many physicists in the absence of a detailed knowledge believe that the two electrons of two-electron atoms under the Coulomb repulsion between the electrons move not together as one particle but as separated particles possessing the two opposite points of the diameter of the orbit around the nucleus. In fact, the two electrons of opposite spin behave like one particle circulating about the nucleus under the rules of quantum mechanics forming two-electron orbitals in helium, beryllium etc. In my paper of 2008, I showed that the positive vibration energy (Ev) described in eV depends on the Ze charge of nucleus as Ev = 16.95Z - 4.1 Of course in the absence of such a vibration energy Ev it is well-known that the ground state energy E described in eV for two orbiting electrons could be given by the Bohr model as E = (-27.21) Z2. So the combination of the energies of the Bohr model and the vibration energies due to the opposite spin of two electrons led to my discovery of the ground state energy of two-electron atoms given by -E = (-27.21) Z2 + (16.95 )Z - 4.1 For example the laboratory measurement of the ionization energy of H- yields an energy of the ground state -E = - 14.35 eV. In this case since Z = 1 we get -E = -27.21 + 16.95 - 4.1 = -14.35 eV. In the same way writing for the helium Z = 2 we get -E = - 108.8 + 32.9 - 4.1 = -79.0 eV The discovery of this simple formula based on the well-established laws of electromagnetism was the first fundamental equation for understanding the energies of many-electron atoms, while various theories based on qualitative symmetry properties lead to complications. Category:Fundamental physics concepts